Call a function with argument list in python

To expand a little on the other answers:

In the line:

def wrapper(func, *args):

The * next to args means "take the rest of the parameters given and put them in a list called args".

In the line:

    func(*args)

The * next to args here means "take this list called args and 'unwrap' it into the rest of the parameters.

So you can do the following:

def wrapper1(func, *args): # with star
    func(*args)

def wrapper2(func, args): # without star
    func(*args)

def func2(x, y, z):
    print x+y+z

wrapper1(func2, 1, 2, 3)
wrapper2(func2, [1, 2, 3])

In wrapper2, the list is passed explicitly, but in both wrappers args contains the list [1,2,3].

The literal answer to your question (to do exactly what you asked, changing only the wrapper, not the functions or the function calls) is simply to alter the line

func(args)

to read

func(*args)

This tells Python to take the list given (in this case, args) and pass its contents to the function as positional arguments.

This trick works on both "sides" of the function call, so a function defined like this:

def func2(*args):
    return sum(args)

would be able to accept as many positional arguments as you throw at it, and place them all into a list called args.

I hope this helps to clarify things a little. Note that this is all possible with dicts/keyword arguments as well, using ** instead of *.

You can use *args and **kwargs syntax for variable length arguments.

What do *args and **kwargs mean?

And from the official python tutorial

http://docs.python.org/dev/tutorial/controlflow.html#more-on-defining-functions

You need to use arguments unpacking..

def wrapper(func, *args):
    func(*args)

def func1(x):
    print(x)

def func2(x, y, z):
    print x+y+z

wrapper(func1, 1)
wrapper(func2, 1, 2, 3)